SATHEE: Indefinite Integration 1 Question 1

Indefinite Integration 1 Question 1

2. If $\int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x = m (π + n)$ , then $m \cdot n$ is equal to

(a) $- \frac{1}{2}$

(b) 1

(d) -1

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Solution:

Let $I = \int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x$

$\begin{aligned} = \int_{0}^{π / 2} \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} + \frac{1}{\sin x}} d x = \int_{0}^{π / 2} \frac{\cos x}{1 + \cos x} d x \\ = \int_{0}^{π / 2} \frac{\cos x (1 - \cos x)}{1 - \cos^{2} x} d x \\ = \int_{0}^{π / 2} \frac{\cos x - \cos^{2} x}{\sin^{2} x} d x \\ = \int_{0}^{π / 2} (cosec x \cot x - \cot^{2} x) d x \\ = \int_{0}^{π / 2} (cosec x \cot x - {cosec}^{2} x + 1) d x \\ = [- cosec x + \cot x + x]_{0}^{π / 2} \\ = x + \frac{\cos x - 1}{\sin x} 0 x + \frac{- 2 \sin^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} 0 \\ = x - \tan \frac{x}{2} = \frac{π}{2} - 1 = \frac{1}{2} [π - 2] \\ = m [π + n] \end{aligned}$

On comparing, we get $m = \frac{1}{2}$ and $n = - 2$

[given]

$∴ m \cdot n = - 1$

(given)

Alternate Solution

Let $I = \int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x$

$= \int_{0}^{π / 2} \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} + \frac{1}{\sin x}} d x$

$= \int_{0}^{π / 2} \frac{\cos x}{\cos x + 1} d x$

$= \int_{0}^{π / 2} \frac{2 \cos^{2} \frac{x}{2} - 1}{2 \cos^{2} \frac{x}{2}} d x$

$∵ \cos θ = 2 \cos^{2} \frac{θ}{2} - 1$ and $\cos θ + 1 = 2 \cos^{2} \frac{θ}{2}$

$= \int_{0}^{π / 2} 1 - \frac{1}{2} \sec^{2} \frac{x}{2} d x$

$= x - \tan {\frac{x}{2}}_{0}^{π / 2} = \frac{π}{2} - 1 = \frac{1}{2} (π - 2)$

Since, $I = m (π - n)$

$∴ m (π - n) = \frac{1}{2} (π - 2)$

On comparing both sides, we get

$m = \frac{1}{2} and n = - 2$

Now, $m n = \frac{1}{2} \times - 2 = - 1$

Indefinite Integration 1 Question 1

2. If $\int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x = m (π + n)$ , then $m \cdot n$ is equal to

Solution:

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Indefinite Integration 1 Question 1

2. If ∫0π/2cot⁡xcot⁡x+cosecxdx=m(π+n), then m⋅n is equal to

Solution:

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2. If $\int_{0}^{π / 2} \frac{\cot x}{\cot x + cosec x} d x = m (π + n)$ , then $m \cdot n$ is equal to